Question

A bottle of wine contains 10.5% ethanol by volume. The density of ethanol (C2H5OH) is 0.789 g/cm3. Calculate the concentration of ethanol in wine in terms of mass percent and molality. mass percent

Answer 1

Answer : The mass percent and molality of ethanol is, 8.28 % and 1.96 mol/kg respectively.

Explanation :

As we are given that 10.5 % ethanol by volume. That means, 10.5 mL of ethanol present in 100 mL of solution.

Density of ethanol = $0.789g/cm^3$

First we have to calculate the mass of ethanol.

$\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}$

$\text{Mass of ethanol}=0.789g/cm^3\times 10.5cm^3=8.28g$

Now we have to calculate the volume of water.

Volume of water = Volume of solution - Volume of ethanol

Volume of water = (100 - 8.28) cm³

Volume of water = 91.7 cm³

Now we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}$

Density of water = 1.00 g/cm³

$\text{Mass of water}=1.00g/cm^3\times 91.72cm^3=91.7g$

Total mass = 8.28g + 91.7g = 99.98 g

Now we have to calculate the mass percent of ethanol.

Mass percent of ethanol = $\frac{\text{Mass of ethanol}}{\text{Total mass}}\times 100=\frac{8.28g}{99.98g}\times 100=8.28\%$

Now we have to calculate the Moles of ethanol.

$\text{Moles of ethanol}=\frac{\text{Given mass ethanol}}{\text{Molar mass ethanol}}=\frac{8.28g}{46g/mol}=0.18mol$

Now we have to calculate the molality of ethanol.

$\text{Molality}=\frac{\text{Moles of ethanol}}{\text{Mass of water in (kg)}}$

$\text{Molality}=\frac{0.18mol}{91.7g}=\frac{0.18mol\times 1000}{91.7kg}=1.96mol/kg$

Therefore, the mass percent and molality of ethanol is, 8.28 % and 1.96 mol/kg respectively.